第77页 - 通城学典课时作业本八年级数学苏科版江苏专版

B

C

$-3$

$解：原式=\frac {x^2-(x-1)(x+1)}{x+1}$

$ =\frac {1}{x+1}$

$解：原式=\frac {(x-2)^2}{(x+2)(x-2)}+\frac {x-2}{x(x+2)}+2$

$ =\frac {x-2}{x+2}+\frac {x-2}{x(x+2)}+2$

$ =\frac {x(x-2)+x-2+2x(x+2)}{x(x+2)}$

$ =\frac {x^2-2x+x-2+2x^2+4x}{x(x+2)}$

$ =\frac {3x^2+3x-2}{x^2+2x}$

$解：原式=\frac {a(a-1)}{(a-1)^2}+\frac 1{a-1}=\frac {a+1}{a-1}$

$当a=2时，原式=\frac {2+1}{2-1}=3$

$ 解：原式=\frac {2n(2n-m)+m(2n+m)+4mn}{(2n-m)(2n+m)}$

$ =\frac {(2n+m)^2}{(2n-m)(2n+m)}$

$ =\frac {2n+m}{2n-m}$

$当\frac mn=\frac 15时，n=5m$

$代入得，原式=\frac {2×5m+m}{2×5m-m}=\frac {11}9$

$ 解：\frac{A}{x-1}-\frac{B}{2-x}$

$=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$

$=\frac{(A+B)x-2A-B}{(x-1)(x-2)}$

$=\frac{2x-6}{(x-1)(x-2)}$

$∴\left\{ \begin{array}{l}{A+B=2}\\ {-2A-B=-6}\ \end{array} \right.$

$解得，\left\{ \begin{array}{l}{A=4}\\ {B=-2}\ \end{array} \right.$